\(\int \frac {(d-c^2 d x^2)^3 (a+b \arcsin (c x))}{x^4} \, dx\) [27]

Optimal result

Integrand size = 25, antiderivative size = 178 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {8}{3} b c^3 d^3 \sqrt {1-c^2 x^2}-\frac {b c d^3 \sqrt {1-c^2 x^2}}{6 x^2}+\frac {1}{9} b c^3 d^3 \left (1-c^2 x^2\right )^{3/2}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}+3 c^4 d^3 x (a+b \arcsin (c x))-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+\frac {17}{6} b c^3 d^3 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \] Output:

8/3*b*c^3*d^3*(-c^2*x^2+1)^(1/2)-1/6*b*c*d^3*(-c^2*x^2+1)^(1/2)/x^2+1/9*b* 
c^3*d^3*(-c^2*x^2+1)^(3/2)-1/3*d^3*(a+b*arcsin(c*x))/x^3+3*c^2*d^3*(a+b*ar 
csin(c*x))/x+3*c^4*d^3*x*(a+b*arcsin(c*x))-1/3*c^6*d^3*x^3*(a+b*arcsin(c*x 
))+17/6*b*c^3*d^3*arctanh((-c^2*x^2+1)^(1/2))
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.91 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {d^3 \left (-6 a+54 a c^2 x^2+54 a c^4 x^4-6 a c^6 x^6-3 b c x \sqrt {1-c^2 x^2}+50 b c^3 x^3 \sqrt {1-c^2 x^2}-2 b c^5 x^5 \sqrt {1-c^2 x^2}-6 b \left (1-9 c^2 x^2-9 c^4 x^4+c^6 x^6\right ) \arcsin (c x)+51 b c^3 x^3 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )\right )}{18 x^3} \] Input:

Integrate[((d - c^2*d*x^2)^3*(a + b*ArcSin[c*x]))/x^4,x]
 

Output:

(d^3*(-6*a + 54*a*c^2*x^2 + 54*a*c^4*x^4 - 6*a*c^6*x^6 - 3*b*c*x*Sqrt[1 - 
c^2*x^2] + 50*b*c^3*x^3*Sqrt[1 - c^2*x^2] - 2*b*c^5*x^5*Sqrt[1 - c^2*x^2] 
- 6*b*(1 - 9*c^2*x^2 - 9*c^4*x^4 + c^6*x^6)*ArcSin[c*x] + 51*b*c^3*x^3*Arc 
Tanh[Sqrt[1 - c^2*x^2]]))/(18*x^3)
 

Rubi [A] (warning: unable to verify)

Time = 0.61 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {5192, 27, 2331, 2124, 27, 1192, 25, 1467, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx\)

\(\Big \downarrow \) 5192

\(\displaystyle -b c \int -\frac {d^3 \left (c^6 x^6-9 c^4 x^4-9 c^2 x^2+1\right )}{3 x^3 \sqrt {1-c^2 x^2}}dx-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} b c d^3 \int \frac {c^6 x^6-9 c^4 x^4-9 c^2 x^2+1}{x^3 \sqrt {1-c^2 x^2}}dx-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 2331

\(\displaystyle \frac {1}{6} b c d^3 \int \frac {c^6 x^6-9 c^4 x^4-9 c^2 x^2+1}{x^4 \sqrt {1-c^2 x^2}}dx^2-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 2124

\(\displaystyle \frac {1}{6} b c d^3 \left (-\int \frac {-2 x^4 c^6+18 x^2 c^4+17 c^2}{2 x^2 \sqrt {1-c^2 x^2}}dx^2-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} b c d^3 \left (-\frac {1}{2} \int \frac {-2 x^4 c^6+18 x^2 c^4+17 c^2}{x^2 \sqrt {1-c^2 x^2}}dx^2-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 1192

\(\displaystyle \frac {1}{6} b c d^3 \left (-\frac {\int -\frac {-2 c^6 x^8-14 c^6 x^4+33 c^6}{1-x^4}d\sqrt {1-c^2 x^2}}{c^4}-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{6} b c d^3 \left (\frac {\int \frac {-2 c^6 x^8-14 c^6 x^4+33 c^6}{1-x^4}d\sqrt {1-c^2 x^2}}{c^4}-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 1467

\(\displaystyle \frac {1}{6} b c d^3 \left (\frac {\int \left (2 x^4 c^6+\frac {17 c^6}{1-x^4}+16 c^6\right )d\sqrt {1-c^2 x^2}}{c^4}-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )-\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{3} c^6 d^3 x^3 (a+b \arcsin (c x))+3 c^4 d^3 x (a+b \arcsin (c x))+\frac {3 c^2 d^3 (a+b \arcsin (c x))}{x}-\frac {d^3 (a+b \arcsin (c x))}{3 x^3}+\frac {1}{6} b c d^3 \left (-\frac {-17 c^6 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )-\frac {2}{3} c^6 x^6-16 c^6 \sqrt {1-c^2 x^2}}{c^4}-\frac {\sqrt {1-c^2 x^2}}{x^2}\right )\)

Input:

Int[((d - c^2*d*x^2)^3*(a + b*ArcSin[c*x]))/x^4,x]
 

Output:

-1/3*(d^3*(a + b*ArcSin[c*x]))/x^3 + (3*c^2*d^3*(a + b*ArcSin[c*x]))/x + 3 
*c^4*d^3*x*(a + b*ArcSin[c*x]) - (c^6*d^3*x^3*(a + b*ArcSin[c*x]))/3 + (b* 
c*d^3*(-(Sqrt[1 - c^2*x^2]/x^2) - ((-2*c^6*x^6)/3 - 16*c^6*Sqrt[1 - c^2*x^ 
2] - 17*c^6*ArcTanh[Sqrt[1 - c^2*x^2]])/c^4))/6
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1192

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^( 
2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + 
 c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && 
IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
 

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2124

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px 
, a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - 
 a*d))), x] + Simp[1/((m + 1)*(b*c - a*d))   Int[(a + b*x)^(m + 1)*(c + d*x 
)^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr 
eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] ||  ! 
ILtQ[n, -1])
 

rule 2331

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2   S 
ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; 
 FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
 

rule 5192

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_) 
^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp[ 
(a + b*ArcSin[c*x])   u, x] - Simp[b*c   Int[SimplifyIntegrand[u/Sqrt[1 - c 
^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0 
] && IGtQ[p, 0]
 

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.89

method	result	size
parts	\(-d^{3} a \left (\frac {c^{6} x^{3}}{3}-3 c^{4} x +\frac {1}{3 x^{3}}-\frac {3 c^{2}}{x}\right )-d^{3} b \,c^{3} \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-3 c x \arcsin \left (c x \right )+\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \arcsin \left (c x \right )}{c x}+\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {17 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {25 \sqrt {-c^{2} x^{2}+1}}{9}\right )\)	\(159\)
derivativedivides	\(c^{3} \left (-d^{3} a \left (\frac {c^{3} x^{3}}{3}-3 c x +\frac {1}{3 c^{3} x^{3}}-\frac {3}{c x}\right )-d^{3} b \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-3 c x \arcsin \left (c x \right )+\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \arcsin \left (c x \right )}{c x}+\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {17 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {25 \sqrt {-c^{2} x^{2}+1}}{9}\right )\right )\)	\(161\)
default	\(c^{3} \left (-d^{3} a \left (\frac {c^{3} x^{3}}{3}-3 c x +\frac {1}{3 c^{3} x^{3}}-\frac {3}{c x}\right )-d^{3} b \left (\frac {c^{3} x^{3} \arcsin \left (c x \right )}{3}-3 c x \arcsin \left (c x \right )+\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {3 \arcsin \left (c x \right )}{c x}+\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {17 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}+\frac {c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{9}-\frac {25 \sqrt {-c^{2} x^{2}+1}}{9}\right )\right )\)	\(161\)

Input:

int((-c^2*d*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x,method=_RETURNVERBOSE)
 

Output:

-d^3*a*(1/3*c^6*x^3-3*c^4*x+1/3/x^3-3*c^2/x)-d^3*b*c^3*(1/3*c^3*x^3*arcsin 
(c*x)-3*c*x*arcsin(c*x)+1/3*arcsin(c*x)/c^3/x^3-3*arcsin(c*x)/c/x+1/6/c^2/ 
x^2*(-c^2*x^2+1)^(1/2)-17/6*arctanh(1/(-c^2*x^2+1)^(1/2))+1/9*c^2*x^2*(-c^ 
2*x^2+1)^(1/2)-25/9*(-c^2*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=-\frac {12 \, a c^{6} d^{3} x^{6} - 108 \, a c^{4} d^{3} x^{4} - 51 \, b c^{3} d^{3} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) + 51 \, b c^{3} d^{3} x^{3} \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) - 108 \, a c^{2} d^{3} x^{2} + 12 \, a d^{3} + 12 \, {\left (b c^{6} d^{3} x^{6} - 9 \, b c^{4} d^{3} x^{4} - 9 \, b c^{2} d^{3} x^{2} + b d^{3}\right )} \arcsin \left (c x\right ) + 2 \, {\left (2 \, b c^{5} d^{3} x^{5} - 50 \, b c^{3} d^{3} x^{3} + 3 \, b c d^{3} x\right )} \sqrt {-c^{2} x^{2} + 1}}{36 \, x^{3}} \] Input:

integrate((-c^2*d*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="fricas")
 

Output:

-1/36*(12*a*c^6*d^3*x^6 - 108*a*c^4*d^3*x^4 - 51*b*c^3*d^3*x^3*log(sqrt(-c 
^2*x^2 + 1) + 1) + 51*b*c^3*d^3*x^3*log(sqrt(-c^2*x^2 + 1) - 1) - 108*a*c^ 
2*d^3*x^2 + 12*a*d^3 + 12*(b*c^6*d^3*x^6 - 9*b*c^4*d^3*x^4 - 9*b*c^2*d^3*x 
^2 + b*d^3)*arcsin(c*x) + 2*(2*b*c^5*d^3*x^5 - 50*b*c^3*d^3*x^3 + 3*b*c*d^ 
3*x)*sqrt(-c^2*x^2 + 1))/x^3
 

Sympy [A] (verification not implemented)

Time = 3.97 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.83 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=- \frac {a c^{6} d^{3} x^{3}}{3} + 3 a c^{4} d^{3} x + \frac {3 a c^{2} d^{3}}{x} - \frac {a d^{3}}{3 x^{3}} + \frac {b c^{7} d^{3} \left (\begin {cases} - \frac {x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c^{2}} - \frac {2 \sqrt {- c^{2} x^{2} + 1}}{3 c^{4}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right )}{3} - \frac {b c^{6} d^{3} x^{3} \operatorname {asin}{\left (c x \right )}}{3} + 3 b c^{4} d^{3} \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) - 3 b c^{3} d^{3} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) + \frac {3 b c^{2} d^{3} \operatorname {asin}{\left (c x \right )}}{x} + \frac {b c d^{3} \left (\begin {cases} - \frac {c^{2} \operatorname {acosh}{\left (\frac {1}{c x} \right )}}{2} + \frac {c}{2 x \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} - \frac {1}{2 c x^{3} \sqrt {-1 + \frac {1}{c^{2} x^{2}}}} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac {i c^{2} \operatorname {asin}{\left (\frac {1}{c x} \right )}}{2} - \frac {i c \sqrt {1 - \frac {1}{c^{2} x^{2}}}}{2 x} & \text {otherwise} \end {cases}\right )}{3} - \frac {b d^{3} \operatorname {asin}{\left (c x \right )}}{3 x^{3}} \] Input:

integrate((-c**2*d*x**2+d)**3*(a+b*asin(c*x))/x**4,x)
 

Output:

-a*c**6*d**3*x**3/3 + 3*a*c**4*d**3*x + 3*a*c**2*d**3/x - a*d**3/(3*x**3) 
+ b*c**7*d**3*Piecewise((-x**2*sqrt(-c**2*x**2 + 1)/(3*c**2) - 2*sqrt(-c** 
2*x**2 + 1)/(3*c**4), Ne(c**2, 0)), (x**4/4, True))/3 - b*c**6*d**3*x**3*a 
sin(c*x)/3 + 3*b*c**4*d**3*Piecewise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c 
**2*x**2 + 1)/c, True)) - 3*b*c**3*d**3*Piecewise((-acosh(1/(c*x)), 1/Abs( 
c**2*x**2) > 1), (I*asin(1/(c*x)), True)) + 3*b*c**2*d**3*asin(c*x)/x + b* 
c*d**3*Piecewise((-c**2*acosh(1/(c*x))/2 + c/(2*x*sqrt(-1 + 1/(c**2*x**2)) 
) - 1/(2*c*x**3*sqrt(-1 + 1/(c**2*x**2))), 1/Abs(c**2*x**2) > 1), (I*c**2* 
asin(1/(c*x))/2 - I*c*sqrt(1 - 1/(c**2*x**2))/(2*x), True))/3 - b*d**3*asi 
n(c*x)/(3*x**3)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.36 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=-\frac {1}{3} \, a c^{6} d^{3} x^{3} - \frac {1}{9} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{6} d^{3} + 3 \, a c^{4} d^{3} x + 3 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b c^{3} d^{3} + 3 \, {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b c^{2} d^{3} - \frac {1}{6} \, {\left ({\left (c^{2} \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\sqrt {-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac {2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b d^{3} + \frac {3 \, a c^{2} d^{3}}{x} - \frac {a d^{3}}{3 \, x^{3}} \] Input:

integrate((-c^2*d*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="maxima")
 

Output:

-1/3*a*c^6*d^3*x^3 - 1/9*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^ 
2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*c^6*d^3 + 3*a*c^4*d^3*x + 3*(c*x*arcsin(c 
*x) + sqrt(-c^2*x^2 + 1))*b*c^3*d^3 + 3*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) 
 + 2/abs(x)) + arcsin(c*x)/x)*b*c^2*d^3 - 1/6*((c^2*log(2*sqrt(-c^2*x^2 + 
1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2)*c + 2*arcsin(c*x)/x^3)*b*d 
^3 + 3*a*c^2*d^3/x - 1/3*a*d^3/x^3
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\text {Timed out} \] Input:

integrate((-c^2*d*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^3}{x^4} \,d x \] Input:

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^3)/x^4,x)
 

Output:

int(((a + b*asin(c*x))*(d - c^2*d*x^2)^3)/x^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.89 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^4} \, dx=\frac {d^{3} \left (-6 \mathit {asin} \left (c x \right ) b \,c^{6} x^{6}+54 \mathit {asin} \left (c x \right ) b \,c^{4} x^{4}+54 \mathit {asin} \left (c x \right ) b \,c^{2} x^{2}-6 \mathit {asin} \left (c x \right ) b -2 \sqrt {-c^{2} x^{2}+1}\, b \,c^{5} x^{5}+50 \sqrt {-c^{2} x^{2}+1}\, b \,c^{3} x^{3}-3 \sqrt {-c^{2} x^{2}+1}\, b c x -51 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (c x \right )}{2}\right )\right ) b \,c^{3} x^{3}-6 a \,c^{6} x^{6}+54 a \,c^{4} x^{4}+54 a \,c^{2} x^{2}-6 a \right )}{18 x^{3}} \] Input:

int((-c^2*d*x^2+d)^3*(a+b*asin(c*x))/x^4,x)
 

Output:

(d**3*( - 6*asin(c*x)*b*c**6*x**6 + 54*asin(c*x)*b*c**4*x**4 + 54*asin(c*x 
)*b*c**2*x**2 - 6*asin(c*x)*b - 2*sqrt( - c**2*x**2 + 1)*b*c**5*x**5 + 50* 
sqrt( - c**2*x**2 + 1)*b*c**3*x**3 - 3*sqrt( - c**2*x**2 + 1)*b*c*x - 51*l 
og(tan(asin(c*x)/2))*b*c**3*x**3 - 6*a*c**6*x**6 + 54*a*c**4*x**4 + 54*a*c 
**2*x**2 - 6*a))/(18*x**3)